1/2x4+1/4x6+1/6x8+...+1/2020x2022
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VN
Võ Ngọc Phương
CTVHS
VIP
6 tháng 9 2023
\(\dfrac{1}{2.4}+\dfrac{1}{4.6}+\dfrac{1}{6.8}+...+\dfrac{1}{40.42}\)
\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{40}-\dfrac{1}{42}\right)\)
\(=\dfrac{1}{2}.\left(\dfrac{1}{2}-\dfrac{1}{42}\right)\)
\(=\dfrac{1}{2}.\dfrac{10}{21}\)
\(=\dfrac{5}{21}\)
\(#Wendy.Dang\)
H9
HT.Phong (9A5)
CTVHS
6 tháng 9 2023
\(\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+\dfrac{1}{6\cdot8}+...+\dfrac{1}{40\cdot42}\)
\(=\dfrac{1}{2}\cdot\left(2\cdot\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+\dfrac{1}{6\cdot8}+...+\dfrac{1}{40\cdot42}\right)\)
\(=\dfrac{1}{2}\cdot\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+...+\dfrac{2}{40\cdot42}\right)\)
\(=\dfrac{1}{2}\cdot\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-...+\dfrac{1}{40}-\dfrac{1}{42}\right)\)
\(=\dfrac{1}{2}\cdot\left(1-\dfrac{1}{42}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{41}{42}\)
\(=\dfrac{41}{84}\)
ML
6 tháng 7 2017
\(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{98.100}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}\)
\(=\frac{49}{100}\)
6 tháng 7 2017
Ta có:
\(A=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+....+\frac{1}{98.100}\)
\(\Rightarrow2A=\frac{2}{2.4}+\frac{2}{4.6}+....+\frac{2}{98.100}\)
\(\Rightarrow2A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+....+\frac{1}{98}-\frac{1}{100}\)
\(\Rightarrow2A=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)
\(\Rightarrow A=\frac{49}{100}\div2=\frac{49}{200}\)
Vậy giá trị của biểu thức là \(\frac{49}{200}\)
19 tháng 3 2017
\(\frac{1}{2x4}\)+ \(\frac{1}{4x6}\)+ ... + \(\frac{1}{98x100}\)= \(\frac{1}{2}\)x(\(\frac{4-2}{2x4}\)+\(\frac{6-4}{4x6}\)+ ... + \(\frac{100-98}{98x100}\))
= \(\frac{1}{2}\)x(\(\frac{1}{2}\)-\(\frac{1}{4}\)+\(\frac{1}{4}\)-\(\frac{1}{8}\)+ ... + \(\frac{1}{98}\)-\(\frac{1}{100}\))
= \(\frac{1}{2}\)x(\(\frac{1}{2}\)-\(\frac{1}{100}\)) = \(\frac{49}{200}\)
16 tháng 3 2016
\(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{96.98}+\frac{1}{98.100}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}\)
\(=\frac{49}{100}\)
NC
8
MH
18 tháng 2 2016
1/2.4 + 1/4.6 + 1/6.8 + ... + 1/96.98 + 1/98.100
= 1/2.(2/2.4 + 2/4.6 + 2/6.8 + ... + 2/96.98 + 2/98.100)
= 1/2.(1/2 - 1/4 + 1/4 - 1/6 + ... + 1/96 - 1/98 + 1/98 - 1/100)
= 1/2.(1/2 - 1/100)
= 1/2.49/100
= 49/200
NH
12
29 tháng 3 2015
Gọi biểu thức trên là A, ta có:
A=1/(2x4) + 1/(4x6) + 1/(6x8) + ... + 1/(96x98) + 1/(98x100)
2A=2/(2x4) + 2/(4x6) + 2/(6x8) + ... + 2/(96x98) + 2/(98x100)
2A=1/2-1/4+1/4-1/6+1/6-1/8+...+1/96-1/98+1/98-1/100
giản ước đi, ta có:
2A=1/2-1/4+1/4-1/6+1/6-1/8+...+1/96-1/98+1/98-1/100
2A=1/2-1/100
2A=49/100
=>A=49/100:2
=>A=49/200
KC
3
28 tháng 3 2016
E=2.(2+2)+4.(2+4)+6.(6+2)+....+98.(98+2).
E=2.2+2.2+2.4+4.4+...+98.98+2.98
E=2.(2+4+6+...+98)+(2.2+4.4+6.6+...+98.98)
E=2.2450+40425.4
E=4900+161700
E=166600
Dấu chấm bằng dấu nhân.
Bạn đúng cho mình đi mình giải thích tại sao :(2.2+4.4+…+98.98)=40425.4
\(\frac{1}{2\times4}+\frac{1}{4\times6}+\frac{1}{6\times8}+...+\frac{1}{2020\times2022}\)
\(=\frac{1}{2}\times\left(\frac{2}{2\times4}+\frac{2}{4\times6}+\frac{2}{6\times8}+...+\frac{2}{2020\times2022}\right)\)
\(=\frac{1}{2}\times\left(\frac{4-2}{2\times4}+\frac{6-4}{4\times6}+\frac{8-6}{6\times8}+...+\frac{2022-2020}{2020\times2022}\right)\)
\(=\frac{1}{2}\times\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2020}-\frac{1}{2022}\right)\)
\(=\frac{1}{2}\times\left(\frac{1}{2}-\frac{1}{2022}\right)\)
\(=\frac{505}{2022}\)
\(\frac{1}{2\times4}+\frac{1}{4\times6}+\frac{1}{6\times8}+...+\frac{1}{2020\times2022}=\frac{1}{2}\times\left(\frac{2}{2\times4}+\frac{2}{4\times6}+\frac{2}{6\times8}+...+\frac{2}{2020\times2022}\right)\)
\(=\frac{1}{2}\times\left(\frac{4-2}{2\times4}+\frac{6-4}{4\times6}+\frac{8-6}{6\times8}+...+\frac{2022-2020}{2020\times2022}\right)\)
\(=\frac{1}{2}\times\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2020}-\frac{1}{2022}\right)\)
\(=\frac{1}{2}\times\left(\frac{1}{2}-\frac{1}{2022}\right)=\frac{505}{2022}\)